Question

if a=√3+√2/√3-√2 and b=√3-√2/√3+√2 find the value of 3(a^2-b^2)​

Answer 1

Here, just we have to substitute the values of 'a' & 'b'. That's it! Come, let us solve.

= a²+b²

= [(√3+√2)/(√3-√2)]² + [(√3-√2)/(√3+√2)]²

= (√3+√2)²/(√3-√2)² + (√3-√2)²/(√3+√2)² [b/z, (a/b)² = a²/b²]

= [(√3)²+2(√3)(√2)+(√2)²]/[(√3)²-2(√3)(√2)+(√2)²] + [(√3)²-2(√3)(√2)+(√2)²]/[(√3)²+2(√3)(√2)+(√2)²] [b/z, (a+b)²=(a²+2ab+b²) & (a-b)²=(a²-2ab+b²)]

= (3+2√6+2)/(3–2√6+2)+(3–2√6+2)/(3+2√6+2)

= (5+2√6)/(5–2√6)+(5–2√6)/(5+2√6)

Now, take LCM,

= [(5+2√6)(5+2√6)+(5–2√6)(5–2√6)]/(5–2√6)(5+2√6)

= [(5+2√6)²+(5–2√6)²]/(5)²-(2√6)²

= {[(5)²+2(5)(2√6)+(2√6)²]+[(5)²-2(5)(2√6)+(2√6)²]}/25–4(√6)(√6)

= [(25+20√6+24)+(25–20√6+24)]/25–24

= (49+20√6+49–20√6)/ 1

= 98/1

= 98

Hence, a²+b²=98

Therefore, 98 is the answer for your question. Thank you.

Answer 2

This problem can be solved in 2 ways.

1st :-

a=3–√+2–√3–√−2–√

=(3–√+2–√)(3–√+2–√)(3–√−2–√)(3–√+2–√)

=(3–√+2–√)23−2

=((3–√)2+(2–√)2+2(3–√)(2–√)

=3+2+26–√

=5+26–√

b=3–√−2–√3–√+2–√

=(3–√−2–√)(3–√−2–√)(3–√+2–√)(3–√−2–√)

=(3–√−2–√)23−2

=((3–√)2+(2–√)2+−2(3–√)(2–√)

=3+2−26–√

=5−26–√

Now

a2+b2

We know that

(x+y)2=x2+y2+2xy

⟹x2+y2=(x+y)2−2xy

If we take x=a and y=b then

a2+b2

=(a+b)2−2ab

=(5+26–√+5−26–√)2−2((5+26–√)(5−26–√))

=(10)2−2(25−24)

Here we use identity